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Question

Evaluate
47C4+3j=050jC3+5k=056kC53k

A
57C4
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B
57C5
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C
56C6
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D
56C5
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Solution

The correct option is A 57C4
Expanding , we get
[47C4+47C3+48C3+49C3+50C3]+[56C53+55C52+54C51+53C50+52C49+51C58]
Now
nCr+nCr+1=n+1Cr+1
Hence
[47C4+47C3+48C3+49C3+50C3
=48C4+48C3+49C3+50C3
:
:
=51C4
Now
56C53+55C52+54C51+53C50+52C49+51C58
=56C3+55C3+54C3+53C3+52C3+51C3
nCr=nCnr
Hence Adding we get
[56C3+55C3+54C3+53C3+52C3+51C3]+51C4
=56C3+56C4
=57C4

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